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JEE Advanced Mix Test 56

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JEE Advanced Mix Test 56
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Weekly Quiz Competition
  • Question 1
    4 / -1

    An air-filled parallel plate capacitor having circular plates has a capacitance of 10 pF. When the radii of the plates are increased two times, the distance between them is halved and if a medium of dielectric constant k is introduced, the capacitance increases 16 times. The value of k is (answer in integer)

    Solution

     

  • Question 2
    4 / -1

    Two parallel identical plates carry equal and opposite charges having a uniform charge of 88.9 μC. Positive plate is fixed on the ceiling of a box and the negative plate has to be suspended. If the area of the plates is 6.35 sq. m and 'm' is the mass of the negative plate, then the value of m in kg, is

    Solution

    Force of attraction between the plates = Weight of the negative plate for it to be suspended

     

  • Question 3
    4 / -1

    A solid sphere of radius R has a charge Q distributed in its volume with a charge density p = kra, where k and a are constants and r is the distance from its centre. If the electric field at r = R/2 is 1/8 times that at r = R, find the value of a.

    Solution

  • Question 4
    4 / -1

    A point source of light S is placed a distance 10 cm in front of the center of a mirror of width 20 cm suspended vertically on a wall. An insect walks with a speed 10 cm/s in front of the mirror along a line parallel to the mirror at a distance 20 cm from it as shown in the figure. Find the maximum time (in seconds) during which the insect can see the image of the source S in the mirror.

    Solution

    Insect can see the image of source S in the mirror, so far as it remains in field of view of image overlapping with the road.

    Shaded portion is the field of view, which overlaps with the road upto length PQ. By geometry we can see that, PQ = 3AB = 60 cm

     

  • Question 5
    4 / -1

    2.616 g of an element (M) is heated with NaOH and NaNO3 to produce Na2MO2 and NH3. Ammonia produced is absorbed in 100 mL of 1 N H2SO4. Excess of the acid is back titrated with NaOH solution and required 80 mL of 0.25 M NaOH solution up to the equivalence point. What is the mass (in ‘g’) of hydrogen gas produced when 20 moles of ‘M’ is treated with excess of NaOH?

    [Atomic mass : Cr = 52; Zn = 65.4; Fe = 56; Cu = 63.5; Ni = 28; H = 1]

    Solution

     

  • Question 6
    4 / -1

    2.27 g of mercuric iodide is added into 100 mL, 0.2 M aqueous solution of KI. If KI is 90% dissociated and potassium tetraiodidomercurate(II) is 80% dissociated, then the osmotic pressure of this solution at 300 K is found to be ‘x’ atm. Calculate the value of ‘100x’ if R = 0.08 L atm mol-1 K-1. [Atomic mass : Hg = 200, I = 127]

    [Assume volume of solution remains constant and formation constant of K2 [HgI4] is very large].

    Solution

     

  • Question 7
    4 / -1

    A new element X forms a compound with chlorine, which contains 0.6873 g of the element per gram of chlorine. Specific heat of the element is 0.05 cal/gK. If the formula of chlorine is XClv, then the value of v is

    Solution

  • Question 8
    4 / -1

    Solution

     

  • Question 9
    4 / -1

    If 2 sin α sin β + 3 cos β + 5 cos α sin β = √38  ∀ α, β ∈ R, then |adj (adj A)| is equal to where 

    Solution

     

  • Question 10
    4 / -1

    Directions For Questions

    Directions: The answer to the following question is a single digit integer ranging from 0 to 9.

    ...view full instructions

    An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is

    Solution

    Number of distinct n-digit numbers which can be formed using digits 2, 5 and 7 = 3n

    We have to find n, so that 3n ≥ 900.

    ⇒ 3n–2 ≥ 100

    ⇒ n – 2 ≥ 5

    ⇒ n ≥ 7

    So, the least value of n is 7.

     

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